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The percentage weight of Zn in white vitriol [ZnSO₄ 7H₂O] is approximately equal

The percentage weight of Zn in white vitriol [ZnSO₄ 7H₂O] is approximately equal to Zn=65, S=32, O=16 and H=1)

Options

(a) 33.65%
(b) 32.56%
(c) 23.65%
(d) 22.65%

Correct Answer:

22.65%

Explanation:

Mol. Wt of ZnSO₄ 7H₂O = 65 + 32 + (4 x 16) + 7 (2 x 1 +16)=287
% Mass of Zinc = 65/287 x 100 = 22.65%

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